Question 586487
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{x\ +\ 1}\ +\ \frac{x\ -\ 2}{3}\ =\ \frac{13}{3(x\ +\ 1)}]

You have three denominators, two of which are factors of the third and those two have no common factors.  Hence, the composite denominator (the one in the RHS of the equation) is your Lowest Common Denominator.  Apply the LCD:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{9}{3(x\ +\ 1)}\ +\ \frac{x^2\ -\ x\ -\ 2}{3(x\ +\ 1)}\ =\ \frac{13}{3(x\ +\ 1)}]

Multiply both sides by the now common denominator:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\ +\ x^2\ -\ x\ -\ 2\ =\ 13]

Collect terms:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 6\ =\ 0]

Solve the factorable quadratic.  Check BOTH roots.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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